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In parallelogram ABCD, side AD is produced to a point E and BE intersects CD at F. Prove that $\triangle ABE \sim \triangle CFB$
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In $\triangle ABE$ and $\triangle CFB$,
$\angle AEB = \angle CBF$
$\angle A = \angle C$
$\therefore \triangle ABE \sim \triangle CFB$_a_1.png)
In $\triangle ABE$ and $\triangle CFB$,
$\angle AEB = \angle CBF$
$\angle A = \angle C$
$\therefore \triangle ABE \sim \triangle CFB$